Brachistochrone with velocity - still a cycloid

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Homework Statement


Brachistochrone problem: if the particle is given an initial velocity [tex]v_0 \neq 0[\tex] , show that the path of minimum time is still a cycloid.

Homework Equations


Conservation of energy:
[tex]\frac{1}{2}mv^2-mgy=\frac{1}{2}mv_0^2[\tex]

The Attempt at a Solution


I know how to start the problem, but in the end have to solve the differential equation:
[tex]\frac{dy}{dx}=\sqrt{\frac{k^2-v_0^2-2gy}{v_0^2+2gy}}[\tex]
which I can't solve. Any ideas and hints would be greatly appreciated!
 

Answers and Replies

  • #2
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Anyone with a decent solution?
 
  • #4
Andrew Mason
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Homework Statement


Brachistochrone problem: if the particle is given an initial velocity [tex]v_0 \neq 0[/tex] , show that the path of minimum time is still a cycloid.

Homework Equations


Conservation of energy:
[tex]\frac{1}{2}mv^2-mgy=\frac{1}{2}mv_0^2[/tex]

The Attempt at a Solution


I know how to start the problem, but in the end have to solve the differential equation:
[tex]\frac{dy}{dx}=\sqrt{\frac{k^2-v_0^2-2gy}{v_0^2+2gy}}[/tex]
which I can't solve. Any ideas and hints would be greatly appreciated!
You have to put at forward slash to close a latex instruction ie: /tex

Think of the initial velocity as the result of a vertical change in position of a height y0 along this path from a starting point of 0. Then you just have:

[tex]\frac{dy}{dx}=\sqrt{\frac{k^2-2gy0-2gy}{2gy0+2gy}}[/tex]

which reduces to:

[tex]\frac{dy}{dx}=\sqrt{\frac{k^2-2g(y0+y)}{2g(y0+y)}}[/tex]

[tex]\frac{dy}{dx}=\sqrt{\frac{k^2-2gy'}{2gy'}}[/tex]

the solution to which is a cycloid.

AM
 

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